11.6: Some Special Types of Equilibria
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Learning Objective
 Identify several special chemical equilibria and construct their K_{a} expressions.
In one sense, all chemical equilibria are treated the same. However, there are several classes of reactions that are noteworthy because of either the identities of the reactants and products, or the form of the K_{eq} expression.
Weak Acids and Bases
In Chapter 12  Acids and Bases, we noted how some acids and bases are strong and some are weak. If an acid or base is strong, it is ionized 100% in H_{2}O. HCl(aq) is an example of a strong acid:
\[HCl(aq)=\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{}(aq)\nonumber \]
However, if an acid or base is weak, it may dissolve in H_{2}O, but does not ionize completely. This means that there is an equilibrium between the unionized acid or base and the ionized form. HC_{2}H_{3}O_{2} is an example of a weak acid:
\[HC_{2}H_{3}O_{2}(aq)\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{}(aq)\nonumber \]
HC_{2}H_{3}O_{2} is soluble in H_{2}O (in fact, it is the acid in vinegar), so the reactant concentration will appear in the equilibrium constant expression. But not all the molecules separate into ions. This is the case for all weak acids and bases.
An acid dissociation constant, K_{a}, is the equilibrium constant for the dissociation of a weak acid into ions. Note the a subscript on the K; it implies that the substance is acting as an acid. The larger K_{a} is, the stronger the acid is. Table \(\PageIndex{1}\)  Acid Dissociation Constants for Some Weak Acids, lists several acid dissociation constants. Keep in mind that they are just equilibrium constants.
Acid  K_{a} 

HC_{2}H_{3}O_{2}  1.8 × 10^{−5} 
HClO_{2}  1.1 × 10^{−2} 
H_{2}PO_{4}^{−}  6.2 × 10^{−8} 
HCN  6.2 × 10^{−10} 
HF  6.3 × 10^{−4} 
HNO_{2}  5.6 × 10^{−4} 
H_{3}PO_{4}  7.5 × 10^{−3} 
Table \(\PageIndex{1}\) Acid Dissociation Constants for Some Weak Acids
Note also that the acid dissociation constant refers to one H^{+} ion coming off of the initial reactant. Thus the acid dissociation constant for H_{3}PO_{4} refers to this equilibrium:
\[H_{3}PO_{4}(aq)\rightleftharpoons H^{+}(aq)+H_{2}PO_{4}^{}(aq)\; \; \; \; \; K_{a}=7.5\times 10^{3}\nonumber \]
The H_{2}PO_{4}^{−} ion, called the dihydrogen phosphate ion, is also a weak acid with its own acid dissociation constant:
\[H_{2}PO_{4}^{}(aq)\rightleftharpoons H^{+}(aq)+HPO_{4}^{2}(aq)\; \; \; \; \; K_{a}=6.2\times 10^{8}\nonumber \]
Thus, for socalled polyprotic acids, each H^{+} ion comes off in sequence; each H^{+} ion that ionizes does so with its own characteristic K_{a}.
Example \(\PageIndex{1}\)
Write the equilibrium equation and the K_{a} expression for HSO_{4}^{−} acting as a weak acid.
Solution
HSO_{4}^{−} acts as a weak acid by separating into an H^{+} ion and an SO_{4}^{2}^{−} ion:
\[HSO_{4}^{}(aq)\rightleftharpoons H^{+}(aq)+SO_{4}^{2}(aq)\nonumber \]
The K_{a} is written just like any other equilibrium constant, in terms of the concentrations of products divided by concentrations of reactants:
\[K_{a}=\frac{[H^{+}][SO_{4}^2]}{[HSO_{4}^{}]}\nonumber \]
Exercise \(\PageIndex{1}\)
Write the equilibrium equation and the K_{a} expression for HPO_{4}^{2}^{−} acting as a weak acid.
 Answer

\[HPO_{4}^{2}(aq)\rightleftharpoons H^{+}(aq)+PO_{4}^{3}(aq)\; \; \; \; \; K_{a}=\frac{[H^{+}][PO_{4}^{3}]}{[HPO_{4}^{2}]}\nonumber \]
The K_{a} is used in equilibrium constant problems just like other equilibrium constants are. However, in some cases, we can simplify the mathematics if the numerical value of the K_{a} is small, much smaller than the concentration of the acid itself. Example 11 illustrates this.
Example \(\PageIndex{2}\)
What is the pH of a 1.00 M solution of HC_{2}H_{3}O_{2}? The K_{a} of HC_{2}H_{3}O_{2} is 1.8 × 10^{−5}.
Solution
This is a twopart problem. We need to determine [H^{+}] and then use the definition of pH to determine the pH of the solution. For the first part, we can use an ICE chart:
HC_{2}H_{3}O_{2}(aq)  H^{+}(g)  +  C_{2}H_{3}O_{2}^{−}(g)  

I  1.00  0  0  
C  −x  +x  +x  
E  1.00 − x  +x  +x 
We now construct the K_{a} expression, substituting the concentrations from the equilibrium row in the ICE chart:
\[K_{a}=\frac{[H^{+}][C_{2}H_{3}O_{2}^{}]}{[HC_{2}H_{3}O_{2}]}=\frac{(x)(x)}{(1.00x)}=1.8\times 10^{5}\nonumber \]
Here is where a useful approximation comes in: at 1.8 × 10^{−5}, HC_{2}H_{3}O_{2} will not ionize very much, so we expect that the value of x will be small. It should be so small that in the denominator of the fraction, the term (1.00 − x) will likely be very close to 1.00. As such, we would introduce very little error if we simply neglect the x in that term, making it equal to 1.00:
(1.00 − x) ≈ 1.00 for small values of x
This simplifies the mathematical expression we need to solve:
\[\frac{(x)(x)}{1.00}=1.8\times 10^{5}\nonumber \]
This is much easier to solve than a more complete quadratic equation. The new equation to solve becomes
x^{2} = 1.8 × 10^{−5}
Taking the square root of both sides,
x = 4.2 × 10^{−3}
Because x is the equilibrium concentration of H^{+} and C_{2}H_{3}O_{2}^{−}, we thus have
[H^{+}] = 4.2 × 10^{−3} M
Notice that we are justified by neglecting the x in the denominator; it truly is small compared to 1.00. Now we can determine the pH of the solution:
pH = −log[H^{+}] = −log(4.2 × 10^{−3}) = 2.38
Exercise \(\PageIndex{2}\)
What is the pH of a 0.500 M solution of HCN? The K_{a} of HCN is 6.2 × 10^{−10}.
 Answer

4.75
Weak bases also have dissociation constants, labeled K_{b} (the b subscript stands for base). However, values of K_{b} are rarely tabulated because there is a simple relationship between the K_{b} of a base and the K_{a} of its conjugate acid:
K_{a} × K_{b} = 1.0 × 10^{−14}
Thus it is simple to calculate the K_{b} of a base from the K_{a} of its conjugate acid.
Example \(\PageIndex{3}\)
What is the value of K_{b} for C_{2}H_{3}O_{2}^{−}, which can accept a proton and act as a base?
Solution
To determine the K_{b} for C_{2}H_{3}O_{2}^{−}, we need to know the K_{a} of its conjugate acid. The conjugate acid of C_{2}H_{3}O_{2}^{−} is HC_{2}H_{3}O_{2}. The K_{a} for HC_{2}H_{3}O_{2} is in Table \(\PageIndex{1}\) "Acid Dissociation Constants for Some Weak Acids" and is 1.8 × 10^{−5}. Using the mathematical relationship between K_{a} and K_{b}:
(1.8 × 10^{−5})K_{b} = 1.0 × 10^{−14}Solving,
\[K_{b}=\frac{1.0\times 10^{14}}{1.8\times 10^{5}}=5.6\times 10^{10}\nonumber \]
Exercise \(\PageIndex{3}\)
What is the value of K_{b} for PO_{4}^{3}^{−}, which can accept a proton and act as a base? The K_{a} for HPO_{4}^{2}^{−} is 2.2 × 10^{−13}.
 Answer

4.5 × 10^{−2}
Autoionization of water
In Chapter 12, "Acids and Bases," we introduced the autoionization of water—the idea that water can act as a proton donor and proton acceptor simultaneously. Because water is not a strong acid (Table 12.5.1  Strong Acids and Bases), it must be a weak acid, which means that its behavior as an acid must be described as an equilibrium. That equilibrium is as follows:
\[H_{2}O(l)+H_{2}O(l)\rightleftharpoons H_{3}O^{+}(aq)+OH^{}(aq)\nonumber \]
The equilibrium constant includes [H_{3}O^{+}] and [OH^{−}] but not [H_{2}O(ℓ)] because it is a pure liquid. Hence the expression does not have any terms in its denominator:
K = [H_{3}O^{+}][OH^{−}] ≡ K_{w} = 1.0 × 10^{−14}
This is the same K_{w} that was introduced in Chapter 12 and the same 1.0 × 10^{−14} that appears in the relationship between the K_{a} and the K_{b} of a conjugate acidbase pair. In fact, we can rewrite this relationship as follows:
K_{a} × K_{b} = K_{w}
Insoluble Compounds
In Chapter 4, section 4.3: "Types of Chemical Reactions  Single and Double Displacement Reactions," the concept of soluble and insoluble compounds was introduced. Solubility rules were presented that allow a person to predict whether certain simple ionic compounds will or will not dissolve.
Describing a substance as soluble or insoluble is a bit misleading because virtually all substances are soluble; they are just soluble to different extents. In particular, for ionic compounds, what we typically describe as an insoluble compound can actually be ever so slightly soluble; an equilibrium is quickly established between the solid compound and the ions that do form in solution. Thus, the hypothetical compound MX does in fact dissolve but only very slightly. That means we can write an equilibrium for it:
\[MX(s)\rightleftharpoons M^{+}(aq)+X^{}(aq)\nonumber \]
The equilibrium constant for a compound normally considered insoluble is called a solubility product constant and is labeled K_{sp} (with the subscript sp, meaning "solubility product"). Because the reactant is a solid, its concentration does not appear in the K_{sp} expression, so like K_{w}, expressions for K_{sp} do not have denominators. For example, the chemical equation and the expression for the K_{sp} for AgCl, normally considered insoluble, are as follows:
\[AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{}(aq)\; \; \; \; \; K_{sp}=[Ag^{+}][Cl^{}]\nonumber \]
Table \(\PageIndex{2}\)  Solubility Product Constants for Slightly Soluble Ionic Compounds, lists some values of the K_{sp} for slightly soluble ionic compounds.
Compound  K_{sp} 

BaSO_{4}  1.1 × 10^{−10} 
Ca(OH)_{2}  5.0 × 10^{−6} 
Ca_{3}(PO_{4})_{2}  2.1 × 10^{−33} 
Mg(OH)_{2}  5.6 × 10^{−12} 
HgI_{2}  2.9 × 10^{−29} 
AgCl  1.8 × 10^{−10} 
AgI  8.5 × 10^{−17} 
Ag_{2}SO_{4}  1.5 × 10^{−5} 
Table \(\PageIndex{2}\) Solubility Product Constants for Slightly Soluble Ionic Compounds.
Example \(\PageIndex{4}\)
Write the K_{sp} expression for Ca_{3}(PO_{4})_{2}.
Solution
Recall that when an ionic compound dissolves, it separates into its individual ions. For Ca_{3}(PO_{4})_{2}, the ionization reaction is as follows:
\[Ca_{3}(PO_{4})_{2}(s)\rightleftharpoons 3Ca^{2+}(aq)+2PO_{4}^{3}(aq)\nonumber \]
Hence the K_{sp} expression is
K_{sp} = [Ca^{2}^{+}]^{3}[PO_{4}^{3}^{−}]^{2}
Exercise \(\PageIndex{4}\)
Write the K_{sp} expression Ag_{2}SO_{4}.
 Answer

K_{sp} = [Ag^{+}]^{2}[SO_{4}^{2}^{−}]
Equilibrium problems involving the K_{sp} can also be done, and they are usually more straightforward than other equilibrium problems because there is no denominator in the K_{sp} expression. Care must be taken, however, in completing the ICE chart and evaluating exponential expressions.
Example \(\PageIndex{5}\)
What are [Ag^{+}] and [Cl^{−}] in a saturated solution of AgCl? The K_{sp} of AgCl is 1.8 × 10^{−10}.
Solution
The chemical equation for the dissolving of AgCl is
\[AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{}(aq)\nonumber \]
The K_{sp} expression is as follows:
K_{sp} = [Ag^{+}][Cl^{−}]So the ICE chart for the equilibrium is as follows:
AgCl(s)  Ag^{+}(aq)  +  Cl^{−}(aq)  

I  0  0  
C  −x  +x  +x  
E  +x  +x 
Notice that we have little in the column under AgCl except the stoichiometry of the change; we do not need to know its initial or equilibrium concentrations because its concentration does not appear in the K_{sp} expression. Substituting the equilibrium values into the expression:
(x)(x) = 1.8 × 10^{−10}
Solving,
x^{2} = 1.8 × 10^{−}^{10}x = 1.3 × 10^{−5}
Thus [Ag^{+}] and [Cl^{−}] are both 1.3 × 10^{−5} M.
Exercise \(\PageIndex{5}\)
What are [Ba^{2}^{+}] and [SO_{4}^{2}^{−}] in a saturated solution of BaSO_{4}? The K_{sp} of BaSO_{4} is 1.1 × 10^{−10}.
 Answer

1.0 × 10^{−5} M
Example \(\PageIndex{6}\)
What are [Ca^{2}^{+}] and [PO_{4}^{3}^{−}] in a saturated solution of Ca_{3}(PO_{4})_{2}? The K_{sp} of Ca_{3}(PO_{4})_{2} is 2.1 × 10^{−33}.
Solution
This is similar to Example 14, but the ICE chart is much different because of the number of ions formed.
Ca_{3}(PO_{4})_{2}(s)  3Ca^{2}^{+}(aq)  +  2PO_{4}^{3}^{−}(aq)  

I  0  0  
C  −x  +3x  +2x  
E  +3x  +2x 
For every unit of Ca_{3}(PO_{4})_{2} that dissolves, three Ca^{2}^{+} ions and two PO_{4}^{3}^{−} ions are formed. The expression for the K_{sp} is also different:
K_{sp} = [Ca^{2}^{+}]^{3}[PO_{4}^{3}^{−}]^{2} = 2.1 × 10^{−33}
Now when we substitute the unknown concentrations into the expression, we get
(3x)^{3}(2x)^{2} = 2.1 × 10^{−33}
When we raise each expression inside parentheses to the proper power, remember that the power affects everything inside the parentheses, including the number. So,
(27x^{3})(4x^{2}) = 2.1 × 10^{−33}
Simplifying,
108x^{5} = 2.1 × 10^{−33}
Dividing both sides of the equation by 108, we get
x^{5} = 1.9 × 10^{−35}
Now we take the fifth root of both sides of the equation (be sure you know how to do this on your calculator):
x = 1.1 × 10^{−7}
We are not done yet. We still need to determine the concentrations of the ions. According to the ICE chart, [Ca^{2}^{+}] is 3x, not x. So
[Ca^{2}^{+}] = 3x = 3 × 1.1 × 10^{−7} = 3.3 × 10^{−7} M
[PO_{4}^{3}^{−}] is 2x, so
[PO_{4}^{3}^{−}] = 2x = 2 × 1.1 × 10^{−7} = 2.2 × 10^{−7} M
Exercise \(\PageIndex{6}\)
What are [Mg^{2}^{+}] and [OH^{−}] in a saturated solution of Mg(OH)_{2}? The K_{sp} of Mg(OH)_{2} is 5.6 × 10^{−12}.
 Answer

[Mg^{2}^{+}] = 1.1 × 10^{−4} M; [OH^{−}] = 2.2 × 10^{−4} M
Food and Drink Application: Solids in Your Wine Bottle
People who drink wine from bottles (as opposed to boxes) will occasionally notice some insoluble materials in the wine, either crusting the bottle, stuck to the cork, or suspended in the liquid wine itself. The accompanying figure shows a cork encrusted with colored crystals. What are these crystals?
One of the acids in wine is tartaric acid (H_{2}C_{4}H_{4}O_{6}). Like the other acids in wine (citric and malic acids, among others), tartaric acid imparts a slight tartness to the wine. Tartaric acid is rather soluble in H_{2}O, dissolving over 130 g of the acid in only 100 g of H_{2}O. However, the potassium salt of singly ionized tartaric acid, potassium hydrogen tartrate (KHC_{4}H_{4}O_{6}; also known as potassium bitartrate and better known in the kitchen as cream of tartar), has a solubility of only 6 g per 100 g of H_{2}O. Thus, over time, wine stored at cool temperatures will slowly precipitate potassium hydrogen tartrate. The crystals precipitate in the wine or grow on the insides of the wine bottle and, if the bottle is stored on its side, on the bottom of the cork. The color of the crystals comes from pigments in the wine; pure potassium hydrogen tartrate is clear in its crystalline form, but in powder form it is white.
The crystals are harmless to ingest; indeed, cream of tartar is used as an ingredient in cooking. However, most wine drinkers do not like to chew their wine, so if tartrate crystals are present in a wine, the wine is usually filtered or decanted to remove the crystals. Tartrate crystals are almost exclusively in red wines; white and rose wines do not have as much tartaric acid in them.
Key Takeaway
 Equilibrium constants exist for certain groups of equilibria, such as weak acids, weak bases, the autoionization of water, and slightly soluble salts.